In He initialization by He et al. [1], we have $y_{l-1} = w_{l-1}^T x_{l-1} + b_{l-1}$ where $y_{l-1}$, $w_{l-1}$, $x_{l-1}$ are vector valued random variables associated with layer $l$.
Here we let $w_{l-1}$ to have a symmetric distribution around zero and $b_{l-1}=0$. With this condition, the authors said that $y_{l-1}$ has mean 0 and has symmetric distribution around zero.
I understand $E[y_{l-1}] = 0$ because $E[y_{l-1}] = \sum_{i}^{} E[w_{(l-1),i} x_{(l-1),i}] = \sum_{i}^{} E[w_{(l-1),i}] E[x_{(l-1),i}] = 0$ using the independence of $w_{l-1}$ and $x_{l-1}$.
But I don't understand why $y_{l-1}$ has the symmetric distribution around zero. How can I show this?
[1] He et.al: http://arxiv.org/abs/1502.01852
For every $y>0$ and $x$, note that \begin{align*} P(Y \geq y|x) &= P(W^{t}x \geq y|x) & \text{W independent of X} \\ &= P((-W^{t})x \leq -y|x) \\ &= P(W^{t}x \leq -y|x) & \text{W is symmetric around 0} \\ &= P(Y \leq -y|x) \end{align*} Conclude that \begin{align*} P(Y \geq y) &= \int_{-\infty}^{\infty}{P(Y \geq y|x)f_{X}(x)dx} \\ &= \int_{-\infty}^{\infty}{P(Y \leq -y|x)f_{X}(x)dx} = P(Y \leq -y) \end{align*} Conclude that $P(Y \geq y) = P(Y \leq -y)$, that is, $Y$ is symmetric around $0$.